Question

Find two consecutive positive integers, sum of whose squares is $365$.

Answer 1

Find the required consecutive integers

Let the two consecutive positive integers be $\left(x\right)$ and $(x+1)$, respectively.

Given, the sum of the square of these two odd positive integers as $365$.

So, ${\left(x\right)}^2+{(x+1)}^2=365$

$$\begin{gathered} \Rightarrow x^2+x^2+1+2x=365 \\ \Rightarrow 2x^2+2x+1-365=0 \\ \Rightarrow 2x^2+2x-364=0 \\ \Rightarrow x^2+x-182=0 \\ \Rightarrow x^2+14x-13x-182=0 \\ \Rightarrow x(x+14)-13(x+14)=0 \\ \Rightarrow (x+14)(x-13)=0 \\ \Rightarrow x=13,-14 \end{gathered}$$

As it is given that the numbers are positive, so $x$ cannot be $-14$.

So, $x=13$ and $(x+1)=(13+1)=14$

Hence, the two consecutive positive odd integers are $13$ and $14$.